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3 years ago

5. Susan exerted 400 newtons of force while pushing on a huge boulder. The boulder moved 0 meters. Calculate work.

Physics

1 answer:

Stells [14]3 years ago

4 0

Answer:

0 J

Explanation:

Work is the product of force and distance in the direction of force.

The formula is ; Work = Force * Distance

Given that :

Force = 400 N

Distance = 0 meters

Work = 400 * 0 = 0 J

No work was done because the boulder did not move.

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Explanation:

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3 years ago

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2. Explain brightness of light using the wave model of light.

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2 years ago

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A ball is dropped from some height. It bounces off the floor and rebounds with a speed that is one-half the speed it had just be

Arada [10]

Answer:

The correct option is C

Explanation:

According to third equation of motion, v

+2ax

Here, u=0 m/s

a=−g and x=−h

Negative sign indicates downward direction. Displacement and acceleration both are downwards.

So,v=±

2(−g)(−h)

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v=−

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′

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2 years ago

An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens.

CaHeK987 [17]

A) 11.1 cm

We can find the focal length of the lens by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p = 16.0 cm is the distance of the object from the lens

q = 36.0 cm is the distance of the image from the lens (taken with positive sign since it is on the opposide side to the image, so it is a real image)

Solving the equation for f:

$\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{36.0 cm}=0.09 cm^{-1}\\f=\frac{1}{0.09 cm^{-1}}=11.1 cm$

B) Converging

The focal length is:

- Positive for a converging lens

- Negative for a diverging lens

In this case, the focal length is positive, so it is a converging lens.

C) 18.0 mm

The magnification equation states that:

$\frac{h_i}{h_o}=-\frac{q}{p}$

where

$h_i$ is the heigth of the image

$h_o$ is the height of the object

$q=36.0 cm$

$p=16.0 cm$

Solving the formula for $h_i$ , we find

$h_i = -h_o \frac{q}{p}=-(8.00 mm)\frac{36.0 cm}{16.0 cm}=-18.0 mm$

So the image is 18 mm high.

D) Inverted

From the magnification equation we have that:

- When the sign of $h_i$ is positive, the image is erect

- When the sign of $h_i$ is negative, the image is inverted

In this case, $h_i$ is negative, so the image is inverted.

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3 years ago